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The cross-sectional area of a pipe in the area of the circle is observed when looking at the pipe from anyone's end. The cross-sectional area formula for the pipe is:

\[Area = \pi (radius)^{2}\]

(where radius corresponds to the radius of the circle as observed from one end of the pipe).

A line segment can be divided into two parts by placing a point on the line in between the two extreme points of the line segment or maybe somewhere externally. In coordinate geometry, section formulas may be used to find a relation between the coordinates of the point which divides the line segment into two parts, and the ratio in which the line segment is divided. If we are provided the coordinates of the point that divides the line segment, we can use the section formula in Mathematics to determine the ratio in which the line segment is divided by the point.

Alternatively, we can figure out the coordinate of the point that divides the line segment in a particular ratio. The section formula may be used to determine the incentre, excentre, and centroid of a triangle. Even in Physics, it is sometimes used to find the centre of mass (COM) of a rigid system of particles.

The section formula is categorized into two types:

Section formula for internal division.

Section formula for external division.

This type of section formula is used when the point divides the line segment internally, that is, the point lies between the two extreme points of the line segment.

Let AB be a line segment where A(x_{1}, y_{1}) and B(x_{2}, y_{2}).

Let P(x, y) be the point which divides the line segment in m:n ratio internally.

By section formula for internal division,

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\[P(x, y) = P(\frac{mx_{2} + nx_{1}}{m+n}, \frac{my_{2} + ny_{1}}{m+n})\]

\[x = (\frac{mx_{2} + nx_{1}}{m+n})\]

\[y = (\frac{my_{2} + ny_{1}}{m+n})\]

Hence, we get the coordinates as P(x, y) which divide the line in the m:n ratio.

Sometimes, m:n is taken to be k:1. In such a case, the formula stands as:

\[P(x, y) = P(\frac{kx_{2} + x_{1}}{k+1}, \frac{ky_{2} + y_{1}}{k+1})\]

\[x = (\frac{kx_{2} + x_{1}}{k+1})\]

\[y = (\frac{ky_{2} + y_{1}}{k + 1})\]

This type of section formula is used when the point divides the line segment externally, that is, the point lies outside the two extreme points of the line segment.

Let AB be a line segment where A(x_{1}, y_{1}) and B(x_{2}, y_{2}).

Let P(x, y) be the point which divides the line segment in m:n ratio externally.

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\[P(x, y) = P(\frac{mx_{2} - nx_{1}}{m - n} , \frac{my_{2} - ny_{1}}{m - n})\]

\[x = (\frac{m_{2} - nx_{1}}{m - n})\]

\[y = (\frac{my_{2} - ny_{1}}{m - n})\]

The section formula is ot only valid in two dimensions but also for coordinate geometry in three-dimensions.

Let AB be a line segment with A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2}).

Let P(x, y, z) be the point which divides the line segment in m:n ratio internally.

\[P(x, y, z) = P(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n}, \frac{mz_{2} + nz_{1}}{m + n})\]

\[x = (\frac{mx_{2} + nx_{1}}{m + n})\]

\[y = (\frac{my_{2} + ny_{1}}{m + n})\]

\[z = (\frac{mz_{2} + nz_{1}}{m + n})\]

When the point P(x, y) divides the line segment into two halves, we may say that P(x, y) is the midpoint of the line segment. By the use of section and midpoint formula:

m : n = 1 : 1, since the line is divided into equal parts, we have.

\[P(x, y) = P(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})\]

\[P(x, y) = P(\frac{(1)x_{2} + (1)x_{1}}{1 + 1}, \frac{(1)y_{2} + (1)y_{1}}{1 + 1})\]

\[P(x, y) = P(\frac{x_{2} + x_{1}}{2}, \frac{y_{2} + y_{1}}{2})\]

∴ \[x = (\frac{x_{2} + x_{1}}{2}), y = (\frac{y_{2} + y_{1}}{2})\]

Q1. Find the coordinates of a point M which divides a line segments PQ in the ratio 2:3. PQ line segment joins the points P (2,1) and Q (-3,6). Does the point M lie on the line 5y - x = 15?

Soln. Given,

P (2, 1), Q (-3, 6),

Let the x coordinate of M be x and y coordinate be y: M (x,y)

m : n = 2 : 3

By section formula,

\[M(x, y) = M(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})\]

\[M(x, y) = M(\frac{2(-3) + 3(2)}{2 + 3}, \frac{2(6) + 3(1)}{2 + 3})\]

\[M(x, y) = M(\frac{-6+6}{5}, \frac{15}{5})\]

M(x, y) = M(0, 3)

x = 0, y = 3

Hence the point M is (0,3).

Putting the values of x and y in 5y - x = 15,

5(3) - (0) - 15 = 0

Hence, point M lies on the line 5y - x = 15.

Q2. Find the midpoint of the line segment AB which joins A (4,8) and (2,4).

Soln. Given,

A (4, 8), B (2, 4)

Let the x coordinate of midpoint be x and y coordinate y: P (x,y).

m : n = 1 : 1

By the use of section and midpoint formula,

\[P(x, y) = P(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})\]

\[P(x, y) = P(\frac{(1)x_{2} + (1)x_{1}}{1 + 1}, \frac{(1)y_{2} + (1)y_{1}}{1 + 1})\]

\[P(x, y) = P(\frac{x_{1} + x_{2}}{2}, \frac{y_{2} + y_{1}}{2})\]

\[P(x, y) = P(\frac{4 + 2}{2}, \frac{8 + 4}{2})\]

P(x, y) = P(3, 6)

∴ x = 3, y = 6

Hence, the midpoint of AB is P (3,6).

FAQ (Frequently Asked Questions)

Q1. What are the Applications of the Section Formula?

Ans. In coordinate geometry, the application of the section formula is huge. It is as important as the distance formula. The section formula may be used to find out the point which segregates a line segment into two parts for a given ratio. Again, it may be used to find the ratio in which a point divides a line segment into two parts. Similarly, the section formula may be used to find out the midpoint of a given line segment. The section formula also helps us find the centroid, excentre, and incentre of a triangle. For example, the centroid formula looks like:

G(x , y) = G[(x_{1} + x_{2} + x_{3})/3 , (y_{1} + y_{2} + y_{3})/3]

The section formula helps solve problems based on geometrical figures like rectangle, square, rhombus, and parallelogram using the inherent properties of these shapes.

Q2. Explain the Points of Trisection.

Ans. Section formula is generally used for a line that is divided into two parts (equal or unequal) by one point. However, we may also use the concept of the section formula for a line segment that is trisected by two points. Trisection means dividing a given line segment into three equal parts. This can be done by placing two points on the line and diving the line into three equal parts. Let AB be a line segment which is divided into three parts by two points P and Q such as AP = PQ = QB. Hence, P and Q are the points of trisection. P divides AB in the ratio 1:2 and Q divides AB in the ratio 2:1.